0

Let $H$ be a separable complex Hilbert space and $T\in\mathcal{B}(H)$. Let $\{v_i\}_{i\in\mathbb{N}}$ be an orthonormal basis for $H$. Then $T$ is of trace class if $$\sum_{i\in\mathbb{N}}\langle |T|v_i,v_i\rangle_H<\infty,$$ where $|T|$ is defined to be $\sqrt{T^*T}$.

I have a couple of very basic questions relating to this definition.

Question 1: Would this definition change if we instead defined $|T|$ to be $\sqrt{TT^*}$?

For a trace class operator, its trace is defined to be $$\text{tr}(T):=\sum_{i\in\mathbb{N}}\langle Tv_i,v_i\rangle_H.$$

Question 2: What is the motivation for defining the notion of trace class using the absolute value, instead of just saying that $T$ is trace class if its trace is finite, i.e. if $$\sum_{i\in\mathbb{N}}\langle Tv_i,v_i\rangle_H<\infty?$$

geometricK
  • 5,115

1 Answers1

2

Q1: I'll assume you already know that the definition does not depend on the choice of the ONB. Let $T=U|T|$ be the polar decomposition of $T$, where $U$ is a partial isometry with initial space $\overline{\mathrm{ran}}(T^\ast)$ and range $\overline{\mathrm{ran}}(T)$. Then $TT^\ast=U|T|^2U^\ast$, hence $|T^\ast|=U|T|U^\ast$.

Now let $(v_j)$ be an ONB of $\overline{\mathrm{ran}}(T)$ and $(w_k)$ an ONB of $\overline{\mathrm{ran}}(T)^\perp$. Then $(U^\ast v_j)$ is an ONB of $\overline{\mathrm{ran}}(T^\ast)=(\ker T)^\perp=(\ker |T|)^\perp$ and $U^\ast w_k=0$. Thus $$ \mathrm{tr}(|T^\ast|)=\sum_j \langle |T^\ast|v_j,v_j\rangle+\sum_k \langle |T^\ast|w_k,w_k\rangle=\sum_j \langle |T|U^\ast v_j,U^\ast v_j\rangle+\sum_k\langle |T|U^\ast w_k,U^\ast w_k\rangle=\mathrm{tr}(|T|). $$ In particular, the left side is finite if and only if the right side is. More generally, the same argument shows that $\mathrm{tr}(f(T^\ast T))=\mathrm{tr}(f(TT^\ast))$ for any postive bounded Borel function $f$.

Q2: The problem is that the terms of this serious don't necessarily have the same sign, so it may diverge without being infinity.

MaoWao
  • 16,366
  • 2
  • 21
  • 43